Take any $n\times k$ matrix $C$ with $r\left(C\right)=k$. The rank condition implies $\left(C^{T}C\right)^{-1}$ exists. Define $M_{C}=I-C\left(C^{T}C\right)^{-1}C^{T}$ 1. For any $C$, how can I show that $M_{C}$ is symmetric and idempotent. What is $r\left(M_{C}\right)$? 2. What is $M_{C}C$? Is there any intuition behind this? 3. Let $A$ be an $n\times k_{A}$ matrix with $r\left(A\right)=k_{A}$. Similarly, let $B$ be an $n\times k_{B}$ with $r\left(B\right)=k_{B}$. Further, suppose $r\left(\left[A,B\right]\right)=k_{1}+k_{2}$. Do $\left(A^{T}M_{B}A\right)^{-1}$ and $\left(B^{T}M_{A}B\right)^{-1}$ exist? 4. Prove that $\left[M_{M_{A}B}\right]A=A$. Is there any intuition behind this?
2025-01-13 00:01:50.1736726510
Take any $n\times k$ matrix $C$ with $r\left(C\right)=k$. $M_{C}=I-C\left(C^{T}C\right)^{-1}C^{T}$
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Indeed, it is idempotent, just follow the definition and you get it: $$ \begin{align} M_{C}^2&=(I-C\left(C^{T}C\right)^{-1}C^{T})(I-C\left(C^{T}C\right)^{-1}C^{T})\\ &=I-C\left(C^{T}C\right)^{-1}C^{T} - C\left(C^{T}C\right)^{-1}C^{T}+C\left(C^{T}C\right)^{-1}C^{T}C\left(C^{T}C\right)^{-1}C^{T}\\ &=I-C\left(C^{T}C\right)^{-1}C^{T} - C\left(C^{T}C\right)^{-1}C^{T}+C\left(C^{T}C\right)^{-1}C^{T}=M_{C} \end{align}$$
and symmetric $$\begin{align} M_{C}^T&= (I-C\left(C^{T}C\right)^{-1}C^{T})^T\\ &=I-(C\left(C^{T}C\right)^{-1}C^{T})^T\\ =&I- (C^{T})^T \left((C^{T}C\right)^{-1})^TC^T\\ =&I- C \left((C^{T}C\right)^T)^{-1}C^T=M_{C} \end{align}$$